The lifespans of bears in a particular zoo are normally distributed. The average bear lives $39.2$ years; the standard deviation is $7.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living less than $17$ years.
Solution: $39.2$ $31.8$ $46.6$ $24.4$ $54$ $17$ $61.4$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $39.2$ years. We know the standard deviation is $7.4$ years, so one standard deviation below the mean is $31.8$ years and one standard deviation above the mean is $46.6$ years. Two standard deviations below the mean is $24.4$ years and two standard deviations above the mean is $54$ years. Three standard deviations below the mean is $17$ years and three standard deviations above the mean is $61.4$ years. We are interested in the probability of a bear living less than $17$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the bears will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the bears will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $17$ years and the other half $({0.15\%})$ will live longer than $61.4$ years. The probability of a particular bear living less than $17$ years is ${0.15\%}$.